There are 10 bulbs in a room,out of which 4 a | vedclass

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(B) Total number of bulbs = $10$. Defective bulbs = $4$. Good bulbs = $6$.We need to choose $3$ bulbs out of $10$. Total ways = $^{10}C_3 = \frac{10 \

Vedclass · Accepted Answer

$29/30$

Vedclass · Answer

(B) Total number of bulbs = $10$. Defective bulbs = $4$. Good bulbs = $6$.We need to choose $3$ bulbs out of $10$. Total ways = $^{10}C_3 = \frac{10 	imes 9 	imes 8}{3 	imes 2 	imes 1} = 120$.The room will be illuminated if at least one bulb is good.It is easier to calculate the complement: the room is $NOT$ illuminated if all $3$ bulbs are defective.Number of ways to choose $3$ defective bulbs = $^4C_3 = 4$.Probability that the room is $NOT$ illuminated = $\frac{4}{120} = \frac{1}{30}$.Probability that the room $IS$ illuminated = $1 - \frac{1}{30} = \frac{29}{30}$.

There are $10$ bulbs in a room,out of which $4$ are defective. If $3$ bulbs are chosen at random,what is the probability that the room will be illuminated? (The room is illuminated if at least one bulb is good.)

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